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Ok,

so this was a recent question I've been asked for homework. "Classify all groups of order $1225$". That's all the question says. How in the world do I approach this?? We have the symmetric group of order $1225$, cyclic group of order $\,1225 = 5*5*7*7\,$ and so we have that as a group.. but the list just seems so big. How do we classify all groups of this order, it's huge! We had a similar question on groups of order $8$ the other week and it wasn't easy..

Thanks for any help guys!

DonAntonio
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Chloe.H
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  • Such questions can very often be addressed using knowledge of the classification of Abelian Groups, and Sylow's theorems to get a handle on possible nonabelian groups. – Mark Bennet Nov 14 '12 at 15:58
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    There's no such a thing as "the symmetric group of order $1225$", as there isn't any natural $n$ such that $n!=1225$. – Kan't Sep 30 '23 at 16:44
  • See also: https://math.stackexchange.com/a/1979758/602386 – Robin May 06 '25 at 09:55

1 Answers1

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Hints:

1) Prove there are exactly one Sylow 5-subgroup and one Sylow 7-subgroup

2) Prove that every group of order the square of a prime number is abelian

3) Show thus that any group of order $\,1225\,$ is abelian and thus

4) There are only $\,4\,$ groups of order $\,1225\,$ (Using the Fundamental Theorem of Finitely Generated Abelian Groups)

DonAntonio
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  • that really helps, thanks a ton! – Chloe.H Nov 19 '12 at 02:17
  • how you concluded from 2 -> 3? if I have 2 groups that are abelian, their multiplication must be also abelian, I don't believe that is true. –  Aug 28 '21 at 09:33
  • @thecorrectanswer I'm not sure what you ask. We have that $;1225 = 25*49;$, and the subgroups of order $;25, 49;$ are normal by (1), Thus, a group of order $;1225;$ is a direct product of two group of order the square of a prime...and we[re done by (2). What isn't clear to you? – DonAntonio Sep 10 '21 at 00:09